Question

Prove that if a and b are odd integers, then 4/(a^2+b^2) (proof by contradiction)

Answer 1

Im actually just looking for the initial negation of the statement.

Answer 2

if a and b are even, then (a^2+b^2) is divisible by 4?

Answer 3

Yes.

Answer 4

But you need to assume the contrary of your initial state.

Answer 5

@Hero

Answer 6

So the negation (contrary) of the statement if a and b are even, then (a^2+b^2) is divisible by 4?

Answer 7

You start with, if a & b are odd, then 4 is divisible by \(a^2+b^2\), was there more to that statement?

Answer 8

*is not if***

Answer 9

Original statement that I am proving is as follows: If a & b are off, then a^2+b^2 is NOT divisible by 4.

Answer 10

I am thinking that the negation of this statement is as follows: If a & b are even, then a^2+b^2 IS divisible by 4.

Answer 11

**odd, not off**

Answer 12

this notation means (a^2+b^2) is NOT divisible by 4

Answer 13

a^2+b^2 mod 4

Answer 14

Proofs are not my strong suit, I think @amistre64 might be better suited to help you.
That said, if your original statement was that they are NOT divisible, then you must assume the contrary, that they ARE.

Answer 15

Okay but do not change the even or off part of the statement?

Answer 16

****Odd

Answer 17

No, because you are trying to prove the contrary of the original condition, which was that odd numbers were not divisible by 4.

Answer 18

okay I will work with that. Thanks

Answer 19

if a and b are odd integers, then 4/(a^2+b^2) ... is that spose to be 4 divides (...)

now contradict it

if a and b are odd integers, then 4\(\not |\) (a^2+b^2)

Answer 20

let a = 2n+1, and b=2k-1 be odd integers for any n,k in \(Z\)

Answer 21

that -1 on the end of b is not going to be product so lets get it to +1 instead
\[a^2+b^2 =(2n+1)^2+(2k+1)^2\]
\[... =(4n^2+4n+1)+(4k^2-4k+1)\]
\[... =4n^2+4n+1+4k^2+4k+1\]
\[... =4(n^2+n+k^2+k)+2\]

Answer 22

lol, forgot to edit it on the second line :) +4k it is

Answer 23

i wonder if completing the squares onm n and k would be useful

Answer 24

\[... =4(n^2+n+k^2+k)+2\]
\[... =4(n^2+n+\frac14-\frac14+k^2+k+\frac14-\frac14)+2\]
\[... =4((n+\frac12)^2+(k+\frac12)^2-\frac12))+2\]
\[... =4((n+\frac12)^2+(k+\frac12)^2)-2+2\]
\[... =4((n+\frac12)^2+(k+\frac12)^2)\]
it should be apparent now that 4 divides it which contradicts the setup

Answer 25

let m = that sum of squares jargon

\[a^2 + b^2 = 4m\]
since 4 divides the right side, 4 also divides the left side